Do this without calculator

First I tried to manipulate the right side like this:

Let $w=\cos\left(\dfrac{2\pi}{7}\right)+i\sin\left(\dfrac{2\pi}{7}\right)$ , so $\dfrac{w^k-\frac{1}{w^k}}{2i}=\sin\left(\dfrac{2\pi k}{7}\right)$ .

The RHS becomes: $\dfrac{w-\frac{1}{w}+w^2-\frac{1}{w^2}+w^3-\frac{1}{w^3}}{2i}$ . Now, using the fact that $\frac{1}{w^k}=w^{7-k}$ then it becomes to: $\dfrac{w-w^6+w^2-w^5+w^3-w^4}{2i}$ .

Now let $a=w+w^2+w^3$ and $b=w^4+w^5+w^6$ , so the RHS is $\dfrac{a-b}{2i}$ . Let's compute $a+b$ and $ab$ using the fact that $\displaystyle\sum_{k=0}^{6} w^k=0$ :

$a+b=w+w^2+w^3+w^4+w^5+w^6=-1 \\ ab=w^5+w^6+1+w^6+1+w+1+w+w^2=3+2(w+\frac{1}{w})+w^2+\frac{1}{w^2}$

With the identity $w^k+\frac{1}{w^k}=2\cos\left(\dfrac{2\pi k}{7}\right)$ we get:

$ab=3+4\cos\left(\dfrac{2\pi}{7}\right)+2\cos\left(\dfrac{4\pi}{7}\right)$

Now, using $(a-b)^2=(a+b)^2-4ab$ we get:

$(a-b)^2=(-1)^2-4\left(3+4\cos\left(\dfrac{2\pi}{7}\right)+2\cos\left(\dfrac{4\pi}{7}\right)\right) \\ (a-b)^2=-16\cos\left(\dfrac{2\pi}{7}\right)-8\cos\left(\dfrac{4\pi}{7}\right)-11$

That is very similar to the LHS, so we choose the positive sign:

$a-b=\sqrt{16\cos\left(\dfrac{2\pi}{7}\right)+8\cos\left(\dfrac{4\pi}{7}\right)+11}i$

Hence, the RHS is $\dfrac{\sqrt{16\cos\left(\dfrac{2\pi}{7}\right)+8\cos\left(\dfrac{4\pi}{7}\right)+11}}{2}$

Comparing both sides we conclude that $x=\boxed{11}$ .

Essentially we want to find

$\left [2 \sin \left ( \frac {2\pi}{7} \right ) + 2 \sin \left ( \frac {4\pi}{7} \right ) + 2 \sin \left ( \frac {6\pi}{7} \right ) \right ]^2 \\ - 16 \cos \left ( \frac {2\pi}{7} \right ) - 8 \cos \left ( \frac {4\pi}{7} \right )$

Expand it:

$4\sin^2 \left ( \frac {2\pi}{7} \right ) + 4\sin^2 \left ( \frac {4\pi}{7} \right ) + 4\sin^2 \left ( \frac {6\pi}{7} \right ) \\ + 8 \sin \left ( \frac {2\pi}{7} \right ) \sin \left ( \frac {4\pi}{7} \right ) + 8 \sin \left ( \frac {2\pi}{7} \right ) \sin \left ( \frac {6\pi}{7} \right ) \\ + 8 \sin \left ( \frac {4\pi}{7} \right ) \sin \left ( \frac {6\pi}{7} \right ) - 16 \cos \left ( \frac {2\pi}{7} \right ) - 8 \cos \left ( \frac {4\pi}{7} \right ) \\$

Apply the trigonometric identities: $2\sin^2 (A) = 1 - \cos (2A), \cos(P) - \cos(Q) = -2 \sin \left ( \frac {P+Q}{2} \right )\sin \left ( \frac {P-Q}{2} \right )$

And for simplicities sake, let $x = \frac {\pi}{7}$ , the expression becomes

$2(1 - \cos(4x)) + 2(1 - \cos(8x) ) + 2 (1 - \cos(12x)) \\ + 4 (-\cos (6x) + \cos(2x)) + 4 ( -\cos(8x) + \cos(4x) ) \\ + 4 ( -\cos(10x) + \cos(2x)) - 16 \cos(2x) - 8 \cos(4x) \\$

Simplifying it and applying the periodic properties of trigonometric functions: $\cos(A) = -\cos(\pi - A), \cos(\pi +A) = -\cos(A)$ gives

$6 + 10 \bigg [\cos(x) - \cos(2x) + \cos(3x) \bigg] \\= 6 + 10 \bigg [ \cos (x) + \cos(3x) + \cos(5x) \bigg ]$

The value of the expression inside the bracket is simply $\frac 1 2$ which could be easily verified . Hence the answer is simply $6 + 10 \left ( \frac 12 \right ) = \boxed{11}$

R.H.S. = 2 S (4) C (2) + S (4) = S (4) [ 2 C (2) + 1]

16 C (2) + 8 C (4) + x = 4 S2 (4) [4 C2 (2) + 4 C (2) + 1]

16 C (2) + 8 C (4) + x = 4 S2 (4) 4 C2 (2) + 4 S2 (4) 4 C (2) + 4 S2 (4)

16 C (2) + 8 C (4) + x = 4 [1 - C (8)] [1 + C (4)] + 4 S2 (4) + 16 S2 (4) C (2)

16 C (2) + 8 C (4) + x = 4 + 4 C (4) - 4 C (8) - 4 C (8) C (4) + 4 S2 (4) + 16 [1 - C2 (4)] C (2)

x = 4 - 4 C (4) - 4 C (8) - 4 C (8) C (4) + 2 [1 - C (8)] - 8 [1 + C (8)] C (2)

x = 4 - 4 C (4) - 4 C (8) - 4 C (8) C (4) + 2 - 2 C (8) - 8 C (2) - 8 C (8) C (2)

x = 6 - 4 C (4) - 6 C (8) - 4 C (8) C (4) - 8 C (2) - 8 C (8) C (2)

x = 6 - 4 c (2) - 6 c (4) - 4 c (4) c (2) - 8 c (1) - 8 c (4) c (1) with domain 1 for 4 Pi/ 7 here.

x = 6 - 4 c (2) - 6 c (4) - 2 [c (6) + c (2)] - 8 c (1) - 4 [c (5) + c (3)]

x = 6 - 4 c (2) - 6 c (4) - 2 c (6) - 2 c (2) - 8 c (1) - 4 c (5) - 4 c (3)

x = 6 - 6 c (2) - 6 c (4) - 2 c (6) - 8 c (1) - 4 c (5) - 4 c (3)

x = 6 - 8 c (1) - 6 c (2) - 4 c (3) - 6 c (4) - 4 c (5) - 2 c (6)

Since c (1) = c (6), c (2) = c (5) and c (3) = c (4) with 1 for 4 Pi/ 7,

x = 6 - 10 c (1) - 10 c (2) - 10 c (3)

x = 6 - 10 [c (1) + c (2) + c (3)]

x = 6 - 10 [-1/ 2]

x = 6 + 5 = 11

I am sorry that this was a result of using calculator.

Cos (4 Pi/ 7) + Cos (8 Pi/ 7) + Cos (12 Pi/ 7) = -0.5 is a special thing which needed to be known.